Anybody solve it?

My Solution, please mention if im wrong. First we have, $T^2\ge 4$ then, $$\sum^n_{k=1} \frac{a \cdot k + \frac{a^2}{4}}{S_k} < T^2 \cdot \sum^n_{k=1} \frac{1}{a_k}= \frac{an(n+1)}{2}\sum_{k=1}^{n}\frac{1}{S_{k}}+\frac{a^2}{4}\sum_{k=1}^{n}\frac{1}{S_k} < 4\sum_{k=1}^{n}\frac{1}{a_k}\Rightarrow \frac{1}{2}\left( an(n+1)\sum_{k=1}^{n}\frac{1}{S_k}+\frac{a^2}{4}\sum_{k=1}^{n}\frac{1}{S_k}\right)< 4\sum_{k=1}^{n}\frac{1}{a_k} $$ \[ \Rightarrow \left( an(n+1)\sum_{k=1}^{n}\frac{1}{S_k}+\frac{a^2}{4}\sum_{k=1}^{n}\frac{1}{S_k}\right)< 8\sum_{k=1}^{n}\frac{1}{a_k} \Rightarrow \left( an(n+1)+\frac{a^2}{2}\right)\left( \sum_{k=1}^{n}\frac{1}{S_k}\right)< 8\sum_{k=1}^{n}\frac{1}{a_k} \] Now we calculate this finite sums. Let's try, \[ \frac{\frac{1}{S_1}+\frac{1}{S_2}+\dots+\frac{1}{S_n}}{\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}}=\frac{\frac{1}{S_1}}{\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}}+\dots+\frac{\frac{1}{S_n}}{\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}} \] Then, \[ \Rightarrow \frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}= \frac{\sum_{sym}\left(\prod_{i=1}^{n-1}a_i\right)}{\prod_{i=1}^{n}a_i}\] \[ \Rightarrow \frac{\prod_{i=1}^{n} a_i}{\left( \sum_{j=0}^{n}(n-j)a_j\right) \left( \sum_{sym} \left(\prod_{i=1}^{n-1}a_i\right)\right)}\left( \frac{(a^2+2an(n+1)}{2}\right)< 16 \] If we choose $n\rightarrow \infty$ then $\frac{\prod_{i=1}^{n} a_i}{\left( \sum_{j=0}^{n}(n-j)a_j\right) \left( \sum_{sym} \left(\prod_{i=1}^{n-1}a_i\right)\right)} \rightarrow 1$ quickly. We assume it has a value equal $1$ also its easy to see $2\le n\le 3$. After trying, solution for $a$ is $1$ and $2$ which is obvious. So we are done !

I can't see in the second row:$$ \Longleftrightarrow \frac{an(n+1)}{2}\sum_{k=1}^{n}\frac{1}{S_{k}}+\frac{a^2}{4}\sum_{k=1}^{n}\frac{1}{S_k}<3\sum_{k=1}^{n}\frac{1}{a_k}$$. For $T^2\ge 4$,why this line is true?

mathmaths wrote: I can't see in the second row:$$ \Longleftrightarrow \frac{an(n+1)}{2}\sum_{k=1}^{n}\frac{1}{S_{k}}+\frac{a^2}{4}\sum_{k=1}^{n}\frac{1}{S_k}<3\sum_{k=1}^{n}\frac{1}{a_k}$$. For $T^2\ge 4$,why this line is true? Cause $$ \frac{an(n+1)}{2}\sum_{k=1}^{n}\frac{1}{S_{k}}+\frac{a^2}{4}\sum_{k=1}^{n}\frac{1}{S_k}= \sum^n_{k=1} \frac{a \cdot k + \frac{a^2}{4}}{S_k} <3 \sum_{k=1}^{n}\frac{1}{a_k}<T^2 \sum_{k=1}^{n}\frac{1}{a_k} \Longleftrightarrow T^2\ge 4>3$$ i think

Is this solution true ?